y = cos x
                                                   e.g. y = (x + 3)2 (x − 1)2
                                  ...
Factorisation of polynomials:                                                           Quadratic formula:                ...
e.g. y = −2e x −1 + 1 , x = −2e y −1 + 1 ,           Differentiation rules:                               The approx. chan...
Anti-differentiation (indefinite integrals):                                  Estimate area by left (or right) rectangles ...
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iTute Notes MM

  1. 1. y = cos x e.g. y = (x + 3)2 (x − 1)2 1  Copyright itute.com 2006 0 π 2π Free download & print from www.itute.com –3 0 1 Do not reproduce by other means –1 Mathematical Methods 3,4 e.g. y = (x + 2 )3 (x − 1) y = tan x Summary sheets Distance between two points –2 0 1 0 π 2π = (x2 − x1 )2 + ( y2 − y1 )2 x +x y +y  Mid-point =  1 2 , 1 2  e.g. y = (x + 2)4  2 2  Modulus functions Parallel lines, m1 = m2  x, x ≥ 0 y= x = Perpendicular lines, − x , x < 0 1 –2 0 m1m2 = −1 or m2 = − Transformations of y = f (x ) m1 Examples of power functions: (1) Vertical dilation (dilation away from the Graphs of polynomial functions in x-axis, dilation parallel to the y-axis) by  1 factorised form: y = x −1  y =  factor k. y = kf (x ) Quadratics e.g. y = (x + 1)(x − 3)  x (2) Horizontal dilation (dilation away from 0 the y-axis, dilation parallel to the x-axis) by 1 factor . y = f (nx ) n –1 0 3 (3) Reflection in the x-axis. y = − f (x ) y = x −2 (4) Reflection in the y-axis. y = f (− x ) e.g. y = (x − 3)2  1  (5) Vertical translation (translation parallel y = 2   to the y-axis) by c units.  x  y = f (x ) ± c , + up, – down. 0 0 3 1 (6) Horizontal translation (translation Cubics e.g. y = 3(x + 1)(x − 1)(x − 2 ) y= x2 parallel to the x-axis) by b units. y = f (x ± b ) , + left, – right. (y = x ) *Always carry out translations last in 0 sketching graphs. Example 1 Sketch y = − 2(x − 1) + 2 Exponential functions: -1 0 1 2 2 y = a x where a = 2, e,10 0 1 2 e.g. y = (x + 1)2 (x − 1) 10x ex 2x –1 0 1 Example 2 Sketch y = 2 1 − x . Rewrite as y = 2 − (x − 1) . 1 2 asymptotic 0 e.g. y = (x + 1) 3 0 1 Logarithmic functions: Relations and functions: –1 0 y = log a x where a = 2, e,10 A relation is a set of ordered pairs (points). If no two ordered pairs have the same first 2x element, then the relation is a function. Quartics e.g. y = (x + 3)(x + 1)(x − 1)(x − 2 ) ex *Use the vertical line test to determine whether a relation is a function. 10x *Use the horizontal line test to determine –3 –1 0 1 2 0 1 whether a function is one-to-one or many-to- asymptotic one. *The inverse of a relation is given by its reflection in the line y = x . *The inverse of a one-to-one function is a e.g. y = (x + 3)2 (x − 1)(x − 2 ) Trigonometric functions: function and is denoted by f −1 . The inverse y = sin x of a many-to-one function is not a function 1 and therefore cannot be called inverse –3 0 1 2 0 π 2π function, and f −1 cannot be used to denote the inverse. –1
  2. 2. Factorisation of polynomials: Quadratic formula: Index laws: ( ) (1) Check for common factors first. 2 Solutions of ax + bx + c = 0 are am n (2) Difference of two squares, a ma n = a m+ n , = am−n , am = a mn an ( ) − 3 = (x − 3)(x + 3) 2 2 − b ± b − 4ac e.g. x 4 − 9 = x 2 2 2 2 x= . 1 1 (ab )n = a nbn , = a−n , am = ( 3 )(x + 3 )(x + 3) 2a = x− 2 n −m Graphs of transformed trig. functions a a (3) Trinomials, by trial and error,  π 1 1 e.g. y = −2 cos 3 x −  + 1 , rewrite a 0 = 1, a 2 =na e.g. 2 x 2 − x − 1 = (2 x + 1)(x − 1)  2 = a,a n (4) Difference of two cubes, e.g.  π ( ) Logarithm laws: equation as y = −2 cos 3 x −  + 1 . x3 − y 3 = (x − y ) x 2 + xy + y 2  6 a log a + log b = log ab, log a − log b = log 3 b (5) Sum of two cubes, e.g. 8 + a = The graph is obtained by reflecting it in the ( 23 + a3 = (2 + a ) 4 − 2a + a 2 ) x-axis, dilating it vertically so that its amplitude becomes 2, dilating it horizontally log ab = b log a, log 1 b = − log b, log a a = 1 (6) Grouping two and two, 2π e.g. x3 + 3x 2 + 3 x + 1 = x3 + 1 + 3 x 2 + 3 x( ) ( ) so that its period becomes 3 , translating log1 = 0, log 0 = undef , log(neg ) = undef ( = (x + 1) x − x + 1 + 3 x(x + 1) 2 ) upwards by 1 and right by π . Change of base: = (x + 1)(x ) 6 log b x 2 − x + 1 + 3x log a x = , log b a = (x + 1)(x+ 2 x + 1 = (x + 1)3 2 ) 3 log e 7 (7) Grouping three and one, e.g. log 2 7 = = 2.8 . log e 2 e.g. x 2 − 2 x − y 2 + 1 π 5π 0 Exponential equations: ( ) = x 2 − 2 x + 1 − y 2 = (x − 1)2 − y 2 –1 6 6 e.g. 2e3 x = 5, e3 x = 2.5 , 3x = loge 2.5 , = (x − 1 − y )(x − 1 + y ) 1 (8) Completing the square, e.g. x= loge 2.5 3 1 1 2 2 Solving trig. equations x2 + x −1 = x2 + x +   −   −1 e.g. 2e 2 x − 3e x − 2 = 0 , 2  2 3 2   2 e.g. Solve sin 2 x = 2 , 0 ≤ x ≤ 2π . ( ) − 3(e )− 2 = 0 , 2 ex 2 x  =  x2   +x+ 1 − 4  5 4  =x +   1  2  −   2 5    ∴ 0 ≤ 2 x ≤ 4π , π 2π π 2π (2e + 1)(e − 2) = 0 , since 2e x x x +1 ≠ 0 ,   2x = , , + 2π , + 2π x x ∴ e − 2 = 0 , e = 2 , x = loge 2 .  3 3 3 3 1 5    x + 1 + 5  = x +  −  π π 7π 4π  2 2  2 2   ∴x = , , , . Equations involving log:    6 3 6 3 e.g. loge (1 − 2 x ) + 1 = 0 , loge (1 − 2 x ) = −1 , (9) Factor theorem, x x e.g. sin = 3 cos , 0 ≤ x ≤ 2π . 1 1 e.g. P(x ) = x3 − 3x 2 + 3 x − 1 2 2 1 − 2 x = e −1 , 2 x = 1 − e −1 , x = 1 −  . 2 e P(− 1) = (− 1)3 − 3(− 1)2 + 3(− 1) − 1 ≠ 0 sin x e.g. log10 (x − 1) = 1 − log10 (2 x − 1) x 2 = 3 , tan x = 3 , P(1) = 13 − 3(1)2 + 3(1) − 1 = 0 0 ≤ ≤π, x log10 (x − 1) + log10 (2 x − 1) = 1 2 2 ∴ (x − 1) is a factor. cos 2 log10 (x − 1)(2 x − 1) = 1 , (x − 1)(2 x − 1) = 10 , Long division: x π 2π ∴ = , ∴x = . 2 x 2 − 3x − 9 = 0 , (2 x + 3)(x − 3) = 0 , x2 − 2x + 1 2 3 3 3 x − 1)x3 − 3 x 2 + 3 x − 1 x = − , 3 . 3 is the only solution because 2 ( − x3 − x 2 ) Exact values for trig. functions: x=− 3 makes the log equation undefined. − 2 x 2 + 3x xo x sin x cos x tan x 2 ( − − 2x2 + 2 x ) 0 30 0 π/6 0 1/2 1 √3/2 0 1/√3 Equation of inverse: x −1 Interchange x and y in the equation to obtain 45 π/4 1/√2 1/√2 1 − (x − 1) 60 π/3 √3/2 1/2 √3 the equation of the inverse. If possible express y in terms of x. 0 90 π/2 1 0 undef ( ∴ P(x ) = (x − 1) x 2 − 2 x + 1 = (x − 1)3 ) 120 2π/3 √3/2 –1/2 –√3 e.g. y = 2(x − 1)2 + 1 , x = 2( y − 1)2 + 1 , x −1 135 3π/4 1/√2 –1/√2 –1 2( y − 1)2 = x − 1 , ( y − 1)2 = , Remainder theorem: 150 5π/6 1/2 –√3/2 –1/√3 2 e.g. when P (x ) = x3 − 3 x 2 + 3 x − 1 is 180 π 0 –1 0 x −1 210 7π/6 –1/2 –√3/2 1/√3 y=± +1 . divided by x + 2 , the remainder is 2 225 5π/4 –1/√2 –1/√2 1 P (− 2) = (− 2)3 − 3(− 2)2 + 3(− 2) − 1 = −11 240 4π/3 –√3/2 –1/2 √3 e.g. y = − 2 +4, x = − 2 +4 , When it is divided by 2 x − 3 , the remainder x −1 y −1 270 3π/2 –1 0 undef 3 1 300 5π/3 –√3/2 1/2 –√3 2 2 is P  = . x−4= − , y −1 = − , 2 8 315 7π/4 –1/√2 1/√2 –1 y −1 x−4 330 11π/6 –1/2 √3/2 –1/√3 2 y=− +1 . 360 2π 0 1 0 x−4
  3. 3. e.g. y = −2e x −1 + 1 , x = −2e y −1 + 1 , Differentiation rules: The approx. change in a function is The product rule: For the multiplication of = f (a + h ) − f (a ) = hf ′(a ) , 1− x 2e y −1 = 1 − x , e y −1 = , two functions, y = u (x )v(x ) , e.g. e.g. find the approx. change in cos x when x 2 y = x 2 sin 2 x , let u = x 2 , v = sin 2 x , π 1− x  1− x  changes from to 1.6. Let f (x ) = cos x , y − 1 = loge   , y = log e   +1. dy du dv 2  2   2  =v +u π dx dx dx then f ′(x ) = − sin x . Let a = , then e.g. y = − loge (1 − 2 x ) − 1 , ( ) = (sin 2 x )(2 x ) + x 2 (2 cos 2 x ) π 2 π x = − loge (1 − 2 y ) − 1 , = 2 x(sin 2 x + x cos 2 x ) f ′(a ) = − sin = −1 and h = 1.6 − = 0.03 2 2 The quotient rule: For the division of loge (1 − 2 y ) = −(x + 1) , 1 − 2 y = e − ( x +1) u (x ) log e x Change in cos x = hf ′(a ) = 0.03×− 1 = −0.03 functions, y = , e.g. y = , 1 2 ( 2 y = 1 − e − ( x +1) , y = 1 − e − ( x +1) .) v(x ) x Rate of change: dy dx is the rate of change of du dv The binomial theorem: v −u dx dy dx dx = y with respect to x. v = , velocity is the e.g. Expand (2 x − 1)4 dx v2 dt = 4C0 (2 x )4 (− 1)0 + 4C1 (2 x )3 (− 1)1 rate of change of position x with respect to (x ) 1  − (loge x )(1)   dv + 4C2 (2 x )2 (− 1)2 + 4C3 (2 x )1 (− 1)3 x 1 − log e x time t. a = , acceleration a is the rate of =   2 = . dt x x2 + 4C4 (2 x )0 (− 1)4 = ...... The chain rule: For composite functions, change of velocity v with respect to t. e.g. Find the coefficient of x2 in the y = f (u ( x) ) , e.g. y = e cos x . Average rate of change: Given y = f (x ) , expansion of (2 x − 3)5 . dy dy du when x = a , y = f (a ) , when x = b , Let u = cos x , y = eu , = × The required term is 5C3 (2 x )2 (− 3)3 dx du dx y = f (b ) , the average rate of change of y ( ) = 10 4 x 2 (− 27 ) = −1080 x 2 . ( )( sin x) = −e = eu − cos x sin x . with respect to x = ∆y = f (b ) − f (a ) . ∴ the coefficient of x2 is –1080. dy ∆x b−a Finding stationary points: Let = 0 and dx Differentiation rules: solve for x and then y, the coordinates of the Deducing the graph of gradient function dy from the graph of a function y = f (x ) = f ' (x ) stationary point. f(x) dx Nature of stationary point at x = a : • ax n anx n −1 Local Local Inflection • max. min. point a(x + c )n an(x + c )n −1 x<a dy dy dy 0 x >0 <0 > 0 , (< 0) a(bx + c )n abn(bx + c )n −1 dx dx dx x=a dy dy dy a sin x a cos x =0 =0 =0 o dx dx dx a sin (x + c ) a cos(x + c ) x>a dy dy dy f’(x) a sin (bx + c ) ab cos(bx + c ) <0 >0 > 0 , (< 0) dx dx dx o o a cos x −a sin x Equation of tangent and normal at x = a : • a cos(x + c ) − a sin (x + c ) 1) Find the y coordinate if it is not given. 0 o x dy a cos(bx + c ) − ab sin (bx + c ) 2) Gradient of tangent mT = at x = a . • dx a tan x a sec 2 x 3) Use y − y1 = mT (x − x1 ) to find equation Deducing the graph of function from the a tan (x + c ) a sec (x + c ) 2 of tangent. graph of anti-derivative function a tan (bx + c ) 1 ab sec 2 (bx + c ) 4) Find gradient of normal mN = − . ∫ f(x)dx+ c mT x x ae ae 5) Use y − y1 = m N (x − x1 ) to find equation • x+c ae ae x + c of the normal. • Linear approximation: 0 x ae bx + c abe bx + c To find the approx. value of a function, use a log e x a f (a + h ) ≈ f (a ) + hf ′(a ) , e.g. find the x approx. value of 25.1 . Let f (x ) = x , o a log e bx a 1 x then f ′(x ) = . Let a = 25 and h = 0.1 , f(x) 2 x o o a log e (x + c ) a then f (a + h ) = 25.1 , f (a ) = 25 = 5 , x+c 0 o • x 1 a log e b(x + c ) a f ′(a ) = = 0.1 . 2 25 • x+c ∴ 25.1 ≈ 5 + 0.1 × 0.1 = 5.01 a log e (bx + c ) ab bx + c
  4. 4. Anti-differentiation (indefinite integrals): Estimate area by left (or right) rectangles Graphics calculator : Pr ( X = a ) = binompdf (n, p, a ) f (x ) ∫ f (x)dx Left Right Pr ( X ≤ a ) = binomcdf (n, p, a ) ax n for n ≠ −1 a n +1 x a b a b Pr ( X < a ) = binomcdf (n, p, a − 1) n +1 Area between two curves: Pr ( X ≥ a ) = 1 − binomcdf (n, p, a − 1) a (x + c )n , n ≠ −1 a (x + c )n +1 y = g (x ) Pr ( X > a ) = 1 − binomcdf (n, p, a ) n +1 y = f (x ) Pr (a ≤ X ≤ b ) = binomcdf (n, p, b ) a (bx + c )n , n ≠ −1 a (bx + c )n +1 a 0 b −binomcdf (n, p, a − 1) (n + 1)b Probability density functions f (x ) for a a log e x , x > 0 x ∈ [a, b] . y y = f (x ) x a log e (− x ) , x < 0 Firstly find the x-coordinates of the a a log e (x + c ) intersecting points, a, b, then evaluate b a c b x x+c a a log e (bx + c ) A= ∫ [ f (x) − g (x)]dx . Always the function a For f (x ) to be a probability density above minus the function below. function, f (x ) > 0 and bx + c b b For three intersecting points: ae x ae x Pr (a < X < b ) = ∫ f (x)dx = 1. a ae x + c ae x + c y = f (x ) c b ae bx + c a bx + c e y = g (x ) Pr ( X < c ) = ∫ f (x)dx , Pr(X > c) = ∫ f (x)dx a c b a b 0 c Normal distributions are continuous prob. a sin x − a cos x distributions. The graph of a normal dist. has b c a sin (x + c ) − a cos(x + c ) ∫ [ f (x) − g (x)]dx ∫ [g (x) − f (x )]dx a bell shape and the area under the graph A= + represents probability. Total area = 1. a sin (bx + c ) a b a − cos(bx + c ) b Discrete probability distributions: ( ) N 1 µ1 , σ 2 , N 2 µ 2 , σ 2 . ( ) In general, in the form of a table, a cos x a sin x x x1 x2 x3 ...... 1 2 µ1 < µ 2 a cos(x + c ) a sin (x + c ) Pr ( X = x ) p1 p2 p3 ...... a cos(bx + c ) a 0 µ1 µ2 X sin (bx + c ) p1 , p2 , p3 ,... have values from 0 to 1 and b p1 + p2 + p3 + ... = 1 . ( 2 ) N1 µ , σ 1 , N 2 µ , σ 2 ( 2 ). Definite integrals: µ = E ( X ) = x1 p1 + x2 p2 + x3 p3 + ... 1 σ1 < σ 2 π π Var ( X ) = x1 p1 + x2 p2 + x3 p3 + ... − µ 2 2 2 2 2  π   π  2 e.g. ∫ 0 2 cos x − dx = sin  x −   3   3  0 σ = sd ( X ) = Var ( X ) 0 X π π   π If random variable Y = aX + b , = sin  −  − sin  0 −  The standard normal distribution: 2 3  3 E (Y ) = aE ( X ) + b , Var (Y ) = a 2 × Var ( X ) has µ = 0 and σ = 1 . N (0,1) and sd (Y ) = a × sd ( X ) . π  π  1+ 3 = sin − sin  −  = . 95% probability interval : (µ − 2δ , µ + 2δ ) µ σ 2 6  3 2 Pr ( A ∩ B ) Properties of definite integrals: Conditional prob: Pr A B = ( ) Pr (B ) . b b 1) ∫ kf (x)dx = k ∫ f (x)dx a a Binomial distributions are examples of discrete prob. distributions. Sampling with 0 Z b b b Graphics calculator: Finding probability, 2) [ f (x ) ± g (x )]dx = f (x )dx ± g (x )dx ∫ ∫ ∫ replacement has a binomial distribution. a a a Number of trials = n. In a single trial, prob. Pr ( X < a ) = normalcdf (− E 99, a, µ , σ ) b c b of success = p, prob. of failure = q = 1- p. Pr ( X > a ) = normalcdf (a, E 99, µ , σ ) ∫ a ∫ ∫ 3) f (x )dx = f (x )dx + f (x )dx , a c The random variable X is the number of successes in the n trials. The binomial dist. Pr (a < X < b ) = normalcdf (a, b, µ , σ ) b a Finding quantile, e.g. given Pr ( X < x ) = 0.7 is Pr ( X = x )= n C x p x q n− x , x = 0,1,2,... with ∫ ∫ where a < c < b . 4) f (x )dx = − f (x )dx a b x = invNorn(0.7, µ , σ ) . b a a µ = np and σ = npq = np(1 − p ) . Given Pr ( X > x ) = 0.7 , then ∫ a ∫ b∫ 4) f (x )dx = − f (x )dx , 5) f (x )dx = 0. a ** Effects of increasing n on the graph of a Pr ( X < x ) = 1 − 0.7 = 0.3 and Area ‘under’ curve: binomial distribution. (1) more points x = invNorm(0.3, µ , σ ) . (2) lower probability for each x value (3) becoming symmetrical , bell shape. X −µ b To find µ and/or σ, use Z = to y = f (x ) A= ∫ f (x)dx a ** Effects of changing p on the graph of a binomial distribution. (1) bell shape when σ convert X to Z first, e.g. find µ given σ = 2 a 0 b p = 0.5 (2) positively skewed if p < 0.5  4−µ y = f (x ) and Pr ( X < 4) = 0.8 . Pr  Z <  = 0.8 , (3) negatively skewed if p > 0.5  2  a c 0 b p = 0.5 p < 0.5 p > 0.5 4−µ c b ∴ = invNorm(0.8) = 0.8416 , 2 ∫ A = − f (x )dx + a ∫ f (x)dx c ∴ µ = 2.3168 .

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